What are the solutions to the equation 0=2x^2-4x+1 ?

The solutions to the equation 0=2x^2-4x+1 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2

Find the zeros of 0=2x^2-4x+1

The zeros of 0=2x^2-4x+1 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2

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0 = 2 x^{2} - 4 x + 1

x = \frac{2 + 2}{2}, x = \frac{2 - 2}{2}

Decimal Notation

x = 1.70710 \dots, x = 0.29289 \dots

Solution steps

0 = 2 x^{2} - 4 x + 1

Switch sides

2 x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 \cdot 1 = 22

x_{1, 2} = \frac{- ( - 4 ) \pm 2 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 2}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 2}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 2}{2 \cdot 2} : \frac{2 + 2}{2}

x = \frac{- ( - 4 ) - 2 2}{2 \cdot 2} : \frac{2 - 2}{2}

The solutions to the quadratic equation are:

x = \frac{2 + 2}{2}, x = \frac{2 - 2}{2}

6 a + 24 = 13 a - 18

2 + 6 - 3

(5 + ∣ x ∣) (11 - x) < 0

\frac{x y}{2 b ^{2}} x \frac{4 b}{1}

200 - 32

What are the solutions to the equation 0=2x^2-4x+1 ?
The solutions to the equation 0=2x^2-4x+1 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2
Find the zeros of 0=2x^2-4x+1
The zeros of 0=2x^2-4x+1 are x=(2+sqrt(2))/2 ,x=(2-sqrt(2))/2